Let $$ f_{\alpha}(x)=\phi_1(\alpha) \mathrm{e}^{-\frac{|x|^\alpha}{\phi_2(\alpha) }}, x \in \mathbb{R}, 0<\alpha<2, $$ where

- $\phi_1(\alpha)=\frac{\alpha}{2} \left\{{\{\Gamma(3/\alpha)\}^{1/2}\over\{ \Gamma(1/\alpha)\}^{3/2}}\right\}$ and
- $\phi_2(\alpha)= \left\{{\Gamma(1/\alpha)\over \Gamma(3/\alpha)}\right\}^{\alpha/2}$.

If $\it{F}_\alpha(\cdot)$ is a Fourier transform of $f_{\alpha}(x)$, show that $\it{F}_{\alpha_1}(\omega) \ge \it{F}_{\alpha_2}(\omega) $, for any $0<\alpha_1\le \alpha_2\le 2$ and $\omega\in \mathbb{R}$.

**UPDATE**. An equivalent problem is to show that Fourier transform of
$$
(\exp(-|x|^\alpha \phi_3(\alpha)))(\phi_1’(\alpha) - |x|^{\alpha}\ln|x|\phi_1(\alpha)\phi_3(\alpha) - |x|^{\alpha}\phi_1(\alpha)\phi’_3(\alpha)),
$$
$$\phi_3(\alpha)= \left\{{\Gamma(3/\alpha)\over \Gamma(1/\alpha)}\right\}^{\alpha/2},
$$
is nonpositive.

**UPDATE TO UPDATE**. Not quite an equivalent but may be somewhat easier problem. Let
$$
f(x,\alpha)=\exp(-|x|B(\alpha))(|x|\ln |x|-A(\alpha)|x|+C(\alpha))
$$
where

- $A(\alpha) = - \frac{\alpha}{2} \ln \left(\frac{\Gamma(3/\alpha)}{\Gamma(1/\alpha)}\right) + \frac{3}{2} \psi_0\left(\frac{3}{\alpha}\right)- \frac{1}{2} \psi_0\left(\frac{1}{\alpha}\right)$
- $B(\alpha)=\left\{{\Gamma(3/\alpha)\over \Gamma(1/\alpha)}\right\}^{\alpha/2}$
- $C(\alpha)= -\frac{1}{\phi_3(\alpha)} \left[1 - \frac{3}{2\alpha}\psi_0\left(\frac{3}{\alpha}\right) +\frac{3}{2\alpha}\psi_0\left(\frac{1}{\alpha}\right)\right]$ where $\psi_0(\cdot)$ is a digamma function.

Prove that Fourier transform of $f(x,\alpha)$ is positive for $0<\alpha<1$ or of $f(|x|^2,\alpha)$ for $0<\alpha<2$

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